WAEC Biology 2011 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 2011 WAEC Biology paper, with answers and explanations. We hold 65 questions from this paper, and the Skolra app has all of them.

WAEC Biology 2011 questions and answers

1. 1.During blood transfusion, agglutination may occur as a result of the reaction between
   1. A
      donor red blood cells and donor plasma proteins
   2. B
      recipient red blood cells and donor white blood cells
   3. C
      donor red blood cell antigens and recipient plasma antibodies
      Correct
   4. D
      recipient plasma antibodies and recipient red blood cell antigens
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   Answer: C

   Agglutination during blood transfusion occurs when the antigens on the donor's red blood cells react with the antibodies in the recipient's plasma. For example, if a person with blood group A (containing anti-B antibodies in plasma) receives blood group B, the anti-B antibodies will agglutinate the B-antigen-bearing red blood cells. This is why blood group compatibility must be carefully checked before transfusion.

2. 2.

   (a) (i) What is the difference between the genotype and phenotype of an individual?
   (ii) State two factors each which determine each of the terms in (a)(i).

   (b) In a breeding experiment, a homozygous dog (BB) was crossed with a homozygous white dog (bb) to produce black puppies in the first filial generation (F1) The second filial generation produced a total of two hundred and forty (240) black and white puppies. With the aid of a genetic diagram, determine:
   (i) the number of white puppies
   (ii) the number of black puppies in the second filial generation.

   1. A
      all black offspring
   2. B
      3 black : 1 white
      Correct
   3. C
      1 black : 1 white
   4. D
      1 black : 2 heterozygous : 1 white
   Show answerHide answer

   Answer: B

   When two heterozygous parents (Bb × Bb) are crossed, the expected genotypic ratio in the offspring is 1BB : 2Bb : 1bb. Since B (black) is dominant over b (white), the phenotypic ratio is 3 black : 1 white. The ratio 1:2:1 describes genotypes, not phenotypes. This principle was established by Gregor Mendel as the Law of Segregation.

3. 3.

   Specimen J (Dry refuse/waste): Paper, dry leaves, or cardboard.

   Specimen K (Wet refuse/waste or sewage): Kitchen waste like vegetable peels, food scraps, or sewage sludge.

   Carefully study specimens J and K and answer questions 4(a) to 4(d).

   (a) Name the type of refuse in specimens: (i) J (ii) K.

   (b) List three animal vectors of disease causing organisms that breed on a each of specimens J and K.

   (ii) Name four diseases each caused by animal vectors associated with each or specimen J and K.

   (c) Name one method of disposal of each of specimens J and K in: (i) rural areas (ii) urban areas.

   (d)(i)State four effects of improper disposal of specimen K

   (ii) In a tabular form. state five differences between specimens J and K.

   (iii) outline one way in which specimen J is of importance to farmers.

   1. A
      Housefly
      Correct
   2. B
      Mosquito
   3. C
      Sandfly
   4. D
      Tsetse fly
   Show answerHide answer

   Answer: A

   The housefly (Musca domestica) breeds in wet refuse, kitchen waste, and sewage, making it a major vector of gastrointestinal diseases such as typhoid, cholera, and dysentery. It transfers pathogens mechanically on its body, feet, and mouthparts to food. Mosquitoes breed in stagnant water rather than refuse, while tsetse flies breed in moist soil and vegetation.

4. 4.Which of the following groups of substances are not six-carbon compounds?
   1. A
      Ribose, deoxyribose, glycerol
      Correct
   2. B
      Glucose, fructose, galactose
   3. C
      Fructose, maltose, starch
   4. D
      Glucose, sucrose, cellulose
   Show answerHide answer

   Answer: A

   Ribose is a five-carbon (pentose) sugar (C₅H₁₀O₅), deoxyribose is also a five-carbon sugar (C₅H₁₀O₄), and glycerol is a three-carbon alcohol (C₃H₈O₃). None of these are six-carbon compounds. Glucose, fructose, and galactose are all hexose monosaccharides with six carbon atoms (C₆H₁₂O₆). Sucrose, cellulose, maltose, and starch are all based on six-carbon glucose units.

5. 5.Vaccination results in
   1. A
      passive immunity through injection of ready-made antibodies
   2. B
      active immunity through stimulation of antibody production
      Correct
   3. C
      destruction of pathogens already present in the body
   4. D
      production of red blood cells to fight infections
   Show answerHide answer

   Answer: B

   Vaccination involves introducing a weakened, killed, or attenuated form of a pathogen (or its antigens) into the body. The immune system recognises these antigens as foreign and produces specific antibodies and memory cells, conferring active immunity. If the person is later exposed to the actual pathogen, memory cells enable a rapid immune response. This differs from passive immunity, which involves administering ready-made antibodies (e.g., antiserum), and provides only temporary protection.

6. 6.Which of the following statements best describes protein synthesis?
   1. A
      Proteins are assembled from amino acids in the nucleus using DNA as a template
   2. B
      DNA is translated directly into protein in the cytoplasm
   3. C
      mRNA carries a copy of the genetic code from DNA to ribosomes where translation occurs
      Correct
   4. D
      tRNA transcribes the genetic code from DNA in the nucleus
   Show answerHide answer

   Answer: C

   Protein synthesis involves two main stages: transcription (DNA → mRNA in the nucleus) and translation (mRNA → protein at ribosomes in the cytoplasm). mRNA (messenger RNA) carries a complementary copy of the genetic code from the DNA in the nucleus to the ribosomes in the cytoplasm, where tRNA brings specific amino acids that are assembled into a polypeptide chain. DNA is never directly translated, and proteins are not assembled in the nucleus.

7. 7.The conversion of glucose to starch in the leaf during the day principally
   1. A
      increases the rate of transpiration
   2. B
      stimulates the opening of stomata
   3. C
      prevents osmotic loss of water from leaf cells
      Correct
   4. D
      provides energy for photosynthesis to continue
   Show answerHide answer

   Answer: C

   Glucose produced during photosynthesis is rapidly converted to starch (an insoluble polysaccharide) in the leaf. This conversion is important because starch does not contribute to the osmotic concentration of the cell sap, preventing osmotic water loss from leaf cells. If glucose accumulated, it would lower the water potential of cell sap and draw in water by osmosis, potentially causing cell damage. Starch also acts as a temporary food store.

8. 8.Which of the following activities promotes forest conservation?
   1. A
      Reforestation and afforestation programmes
      Correct
   2. B
      Slash-and-burn farming
   3. C
      Selective logging of mature trees only
   4. D
      Conversion of forest land to agricultural land
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   Answer: A

   Reforestation (replanting trees in deforested areas) and afforestation (planting trees in areas where there was no previous forest cover) actively promote forest conservation by increasing tree cover, restoring biodiversity, and maintaining ecosystem services. Slash-and-burn farming and conversion of forests to farmland directly destroy forest habitats. Selective logging, while better than clear-cutting, still reduces biodiversity and can cause soil erosion if not carefully managed.

9. 9.A xerophyte conserves water by possession of the following features except
   1. A
      sunken stomata in pits
   2. B
      leaves reduced to spines or needles
   3. C
      large, thin, broad leaves with many stomata
      Correct
   4. D
      thick, waxy cuticle on leaf surface
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   Answer: C

   Xerophytes are plants adapted to survive in arid environments where water is scarce. Their adaptations include sunken stomata (reduce air movement and humidity gradient), thick waxy cuticle (reduces cuticular transpiration), and leaves reduced to spines (minimise surface area for transpiration). Having large, thin, broad leaves with many stomata would greatly increase transpiration, making it a hydrophyte (water plant) adaptation, not a xerophyte adaptation.

10. 10.The removal of all the phloem tissues of the stem of a plant close to the root system for a long period of time is likely to
    1. A
       stop water uptake by the roots
    2. B
       cause the roots to die from lack of organic food
       Correct
    3. C
       increase the rate of transpiration in the leaves
    4. D
       stimulate greater production of xylem tissue
    Show answerHide answer

    Answer: B

    The phloem transports organic nutrients (primarily sucrose) produced by photosynthesis in the leaves downward to all non-photosynthetic parts of the plant, including the roots. If the phloem near the root is removed (ring-barking), the roots are cut off from their supply of organic food and will eventually starve and die. This demonstrates the role of phloem in translocation and the dependence of roots on shoot-produced photosynthates.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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